To find the magnitude of the two charges, we again use Coulomb's Law:

$F = \frac{k \cdot q^2}{r^2}$

Given:

• The force $F = 90 \, \text{N}$
• The separation $r = 0.01 \, \text{m}$
• Coulomb's constant $k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$

Since the charges are identical, let $q_1 = q_2 = q$. Thus, the equation becomes:

$90 = \frac{8.99 \times 10^9 \cdot q^2}{(0.01)^2}$

First, solve for $q^2$:

$90 = \frac{8.99 \times 10^9 \cdot q^2}{0.0001}$

$90 = 8.99 \times 10^{13} \cdot q^2$

$q^2 = \frac{90}{8.99 \times 10^{13}}$

$q^2 \approx 1.001 \times 10^{-12}$

Finally, solve for $q$:

$q \approx \sqrt{1.001 \times 10^{-12}}$

$q \approx 1.001 \times 10^{-6} \, \text{C}$

So, the magnitude of each charge is approximately $1.001 \, \mu\text{C}$.

Would you like more details or have any questions?

Relative Questions:

1. How does changing the distance between charges affect the force according to Coulomb's Law?
2. What is the relationship between force and charge in Coulomb's Law?
3. If the separation distance were doubled, how would the force change?
4. What are the limitations of Coulomb's Law in practical applications?
5. How can you experimentally determine the value of a charge?
6. What are the units of Coulomb's constant $k$?
7. How would the force change if one charge was three times larger than the other?
8. What is the principle of superposition in electrostatics?

Tip:

When working with electrostatic problems, always double-check the units and conversion factors to ensure accuracy in your calculations.